What is the variance of the following data? If necessary, round your answer to two decimal places 600,470,170,430,300

Mathematics

1 answer:

Mkey [24]3 years ago

3 0

Answer:

72112885110079700000000000000.00

Step-by-step explanation:

X(mean) = (600470170430300 + 23056 + 22123 + 21704 + 20365) / 5 = 120094034103510

n=5

S²(variance) = ((600470170430300-120094034103510)²+(23056-120094034103510)²+(22123-120094034103510)²*+(21704-120094034103510)²+(20365-120094034103510)²/(5-1)

S² = 72112885110079700000000000000.00

You might be interested in

Randi scored r points at a basketball game.

steposvetlana [31]

Answer:

b,f

Step-by-step explanation:

b: if sandi scored 3 more points than randi, then sandis points are equal to 3+randi

f: so the equation is 3+r=s,

so input values

3+r=18

r=18-3

r=15

6 0

3 years ago

A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It

lapo4ka [179]

Answer:

A. As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.

B. t=-3.144

C. The critical value for a significance level of 0.05, a two tailed test with 4 degrees of freedom is t=±2.064.

D. P-value = 0.004

E. Reject H0

F. The probability of making a Type I error is equal to the significance level: P(Type I error) = 0.05.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the two processes yield different average errors.

As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

being μ1: average error for Process A and μ2: average error fo Process B.

The significance level is 0.05.

The sample 1, of size n1=12 has a mean of 2 and a standard deviation of 1.

The sample 1, of size n1=14 has a mean of 3 and a standard deviation of 0.5.

The difference between sample means is Md=-1.

$M_d=M_1-M_2=2-3=-1$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1^2}{12}+\dfrac{0.5^2}{14}}\\\\\\s_{M_d}=\sqrt{0.083+0.018}=\sqrt{0.101}=0.318$